MHT CET · Maths · Differentiation
If \(x=2 \cos t-\cos 2 t, \quad y=2 \sin t-\sin 2 t\), then
the value of \(\left.\frac{d^{2} y}{d x^{2}}\right|_{t=\pi / 2}\) is
- A \(3 / 2\)
- B \(5 / 2\)
- C \(5 / 2\)
- D \(-3 / 2\)
Answer & Solution
Correct Answer
(D) \(-3 / 2\)
Step-by-step Solution
Detailed explanation
\(\frac{d x}{d t}=-2 \sin t+2 \sin 2 t\)
\(\frac{d y}{d t} =2 \cos t-2 \cos 2 t \)
\( \therefore \quad \frac{d y}{d x} =\frac{2 \cos t-2 \cos 2 t}{-2 \sin t+2 \sin 2 t} \)
\( =\frac{\cos t-\cos 2 t}{\sin 2 t-\sin t} \)
\( =\frac{2 \sin \frac{3 t}{2} \cdot \sin \frac{t}{2}}{2 \cos \frac{3 t}{2} \cdot \sin \frac{t}{2}} \)
\( =\tan \frac{3 t}{2}\) \(\frac{d^{2} y}{d x^{2}} =\sec ^{2} \frac{3 t}{2} \cdot \frac{3}{2} \cdot \frac{d t}{d x} \)
\( =\frac{3}{2} \sec ^{2} \frac{3 t}{2} \frac{1}{(2 \sin 2 t-2 \sin t)} \)
\( \Rightarrow \quad \left.\frac{d^{2} y}{d x^{2}}\right|_{t=\pi / 2}=-3 / 2\)
\(\frac{d y}{d t} =2 \cos t-2 \cos 2 t \)
\( \therefore \quad \frac{d y}{d x} =\frac{2 \cos t-2 \cos 2 t}{-2 \sin t+2 \sin 2 t} \)
\( =\frac{\cos t-\cos 2 t}{\sin 2 t-\sin t} \)
\( =\frac{2 \sin \frac{3 t}{2} \cdot \sin \frac{t}{2}}{2 \cos \frac{3 t}{2} \cdot \sin \frac{t}{2}} \)
\( =\tan \frac{3 t}{2}\) \(\frac{d^{2} y}{d x^{2}} =\sec ^{2} \frac{3 t}{2} \cdot \frac{3}{2} \cdot \frac{d t}{d x} \)
\( =\frac{3}{2} \sec ^{2} \frac{3 t}{2} \frac{1}{(2 \sin 2 t-2 \sin t)} \)
\( \Rightarrow \quad \left.\frac{d^{2} y}{d x^{2}}\right|_{t=\pi / 2}=-3 / 2\)
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